Structure of Atom — NEET UG practice

1. The principal quantum number ($n$) gives the shell number and appears as the first digit of the orbital name, while the azimuthal quantum number ($l$) decides the subshell letter using the code:

$$l=0 \to s,\quad l=1 \to p,\quad l=2 \to d,\quad l=3 \to f$$

2. For entry A, $n=2$ and $l=1$, so combining these values gives the orbital

$$2p$$

This matches II.

3. For entry B, $n=4$ and $l=0$, so the orbital is

$$4s$$

This matches III.

4. For entry C, $n=5$ and $l=3$, so the orbital is

$$5f$$

This matches IV.

5. For entry D, $n=3$ and $l=2$, so the orbital is

$$3d$$

This matches I.

Hence, the correct matching is A-II, B-III, C-IV, D-I, so the answer is C.

1. Since power is energy delivered per second, the total energy radiated by the bulb in one second equals its power rating.

$$E_{total} = 150 \times 1 = 150\ \text{J}$$

2. Only $8\%$ of this total energy is converted into light, so the light energy produced per second is

$$E_{light} = \frac{150 \times 8}{100} = 12\ \text{J}$$

3. Since the total light energy equals the number of photons multiplied by the energy of a single photon,

$$E_{light} = n \times E_{photon}$$

Therefore, solving for $n$,

$$n = \frac{12}{4.42 \times 10^{-19}} \approx 2.71 \times 10^{19}$$

Hence, the answer is A.

1. Radial nodes are the points (other than $r=0$ and $r=\infty$) where the radial wavefunction $\psi_r$ becomes zero, so the curve crosses the $r$-axis at each radial node.

Given the formula that connects radial nodes to the quantum numbers,

$$\text{Number of radial nodes} = n - l - 1$$

2. Therefore, an orbital having exactly two radial nodes must satisfy

$$n - l - 1 = 2$$

3. Among the four schematic plots, the correct one is the curve that crosses the $r$-axis exactly twice before finally decaying to zero at large $r$, since each crossing represents one radial node.

Hence, the plot satisfying this condition corresponds to option D (whose content is figure "C"), so the answer is D.